\(A=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+.........+\frac{3}{98.101}\right)\)

\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+............+\frac{1}{98}-\frac{1}{101}\right)\)

\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)

\(=\frac{4}{3}.\frac{99}{202}\)

\(=\frac{66}{101}\)

\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{98.101}\) 

\(\frac{4}{3}A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{98.101}\)

\(\frac{4}{3}A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{101}\) 

\(A=\left(\frac{1}{2}-\frac{1}{101}\right).\frac{3}{4}\) 

\(A=\frac{99}{202}.\frac{3}{4}=\frac{297}{808}\)

\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{98.101}\)

\(\Rightarrow A=4\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+....+\frac{1}{98.101}\right)\)

\(\Rightarrow A=4\left[\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{98}-\frac{1}{101}\right)\right]\)

\(\Rightarrow A=4\left[\frac{1}{3}\left(\frac{1}{2}-\frac{1}{101}\right)\right]\Rightarrow A=4\left(\frac{1}{3}.\frac{99}{202}\right)\Rightarrow A=4.\frac{33}{202}\)\(\Rightarrow A=\frac{66}{101}\)

#)Giải :

\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{98.101}\)

\(\Rightarrow3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{99.101}\)

\(\Rightarrow3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\)

\(\Rightarrow3A=\frac{1}{2}-\frac{1}{101}\)

\(\Rightarrow3A=\frac{99}{202}\)

\(\Leftrightarrow A=\frac{33}{202}\)

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{101}\right)\)

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)

\(A=\frac{1}{3}.\frac{99}{202}=\frac{33}{202}\)

\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{65.68}\)

\(A=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{65.68}\right)\)

\(A=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{65}-\frac{1}{68}\right)\)

\(A=\frac{4}{3}.\left[\frac{1}{2}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+...+\left(\frac{1}{65}-\frac{1}{65}\right)-\frac{1}{68}\right]\)

\(A=\frac{4}{3}.\left[\frac{1}{2}-\frac{1}{68}\right]\)

\(A=\frac{4}{3}.\frac{33}{68}\)

\(A=\frac{11}{17}\)

~ Hok tốt ~

\(A=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{65}-\frac{1}{68}\right)\)

     \(=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{68}\right)\)

       \(=\frac{4}{3}\times\frac{33}{68}=\frac{11}{17}\)

S = 4/2.5 + 4/5.8 + 4/8.11 + ... + 4/65.48

S = 4/3 . ( 3/2.5 + 3/5.8 + 3/8.11 + ... + 3/65.68 )

S = 4/3 . ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/65 - 1/68 )

S = 4/3 . ( 1/2 - 1/68 )

S = 4/3 . 33/68

S = 11/17

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(=\frac{11}{17}\)

a) \(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)

\(=4.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\right)\)

\(=4.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=4.\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=4.\frac{32}{99}\)

\(=\frac{128}{99}\)

\(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)

\(=2\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=2.\frac{32}{99}\)

\(=\frac{64}{99}\)

\(\frac{18}{2.5}+\frac{18}{5.8}+...+\frac{18}{203.206}\)

\(=6\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{203.206}\right)\)

\(=6\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{203}-\frac{1}{206}\right)\)

\(=6\left(\frac{1}{2}-\frac{1}{206}\right)\)

\(=6.\frac{102}{206}\)

\(=\frac{612}{206}\) ( tự rút gọn

\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+.........+\frac{4}{65.68}\)

\(A=4\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+.........+\frac{1}{65.68}\right)\)

\(A=\frac{4}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+..........+\frac{3}{65.68}\right)\)

\(A=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-...........-\frac{1}{65}-\frac{1}{68}\right)\)

\(A=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{68}\right)\)

\(A=\frac{4}{3}\left(\frac{34}{68}-\frac{1}{68}\right)\)

\(A=\frac{4}{3}\left(\frac{33}{68}\right)\)

\(A=\frac{11}{17}\)

Vậy A = \(\frac{11}{17}\)

Chúc bạn học tốt!

\(\frac{4}{2\cdot5}\)+\(\frac{4}{5\cdot8}\)+\(\frac{4}{8\cdot11}\)+.......+\(\frac{4}{8\cdot83}\)=\(\frac{4}{3}\) (\(\frac{3}{2\cdot5}\)+\(\frac{3}{5\cdot8}\) +......+\(\frac{3}{80\cdot83}\) )

=\(\frac{4}{3}\) (\(\frac{1}{2}\) -\(\frac{1}{5}\) +\(\frac{1}{5}\) -\(\frac{1}{8}\) +..........+\(\frac{1}{80}\) -\(\frac{1}{83}\) )

=\(\frac{4}{3}\) (\(\frac{1}{2}\) -\(\frac{1}{83}\) )

=\(\frac{4}{3}\)*\(\frac{81}{166}\) 

=\(\frac{54}{83}\)

Gọi biểu thức đó là A

Ta có: \(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{17.20}\)

\(A:4.3=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)

\(A:4.3=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)

\(A:4.3=\frac{1}{2}-\frac{1}{20}\)

\(A:4.3=\frac{9}{20}\)

\(A=\frac{3}{5}\)

k mình nha

tại sao fải chia 4 nhân 3 tui ko hiẻu